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\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{复变函数测验4解答}
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\date{2024 年 5 月 27 日}
%\date{March 9, 2021}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item  %Problem 1 第116页例子4.5
求幂级数 $\sum\limits_{n=1}^{\infty} \sin(in)z^n$ 的收敛半径。

\vspace{0.2cm}

{\color{red}解答：
通项系数 $c_n=\sin(in) = \frac{e^{iin}-e^{-iin}}{2i} = \frac{e^{-n}-e^n}{2i}$, 根据达朗贝尔公式， 
$$
\ell = \lim\limits_{n\to\infty} \left\vert \frac{c_{n+1}}{c_n} \right\vert 
= \lim\limits_{n\to\infty} \left\vert \frac{\sin(i(n+1))}{\sin(in)} \right\vert 
= \lim\limits_{n\to\infty} \left\vert \frac{(e^{-n-1}-e^{n+1})/(2i)}{(e^{-n}-e^n)/(2i)} \right\vert 
=e.
$$
因此该幂级数的收敛半径为 $R=\frac{1}{\ell} = \frac{1}{e}$. 

}

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\item  %Problem 2 第122页例子4.11
求多值解析函数 $f(z)=\sqrt{z-2i}$ 使得 $f(0)=i-1$ 的解析分支在 $z=0$ 处的泰勒展开，写出前四项。

\vspace{0.2cm}

{\color{red}解答：
按照一般的幂函数的定义，
\begin{eqnarray*}
(1+z)^\alpha = e^{\alpha\mathrm{Ln}(1+z)}. 
\end{eqnarray*}
对数函数 $\mathrm{Ln}(1+z)$ 的所有解析分支在 $z=0$ 处的泰勒展式为 
\begin{eqnarray*}
\mathrm{Ln}(1+z) = 2k\pi i +z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots, |z|<1, k=0,\pm 1, \pm 2, \cdots. 
\end{eqnarray*}
由此可以确定 $(1+z)^\alpha$ 的所有解析分支的泰勒展式，
%\begin{eqnarray*}
\begin{subequations}
\begin{align}
(1+z)^\alpha &= \exp\left[ \alpha \left( 2k\pi i + z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots \right) \right] \nonumber\\ 
%, |z|<1, k=0,\pm 1, \pm 2, \cdots. 
&= e^{2k\pi \alpha i} \exp \left[ \alpha \left( z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots \right) \right] \nonumber \\ 
&= e^{2k\pi \alpha i} \left[ 1+\alpha \left( z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots \right) + 
\frac{1}{2!} \alpha^2 \left( z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots \right)^2 + \cdots \right] \nonumber\\ 
&= e^{2k\pi \alpha i} \left[ 1+ \alpha z + \frac{\alpha(\alpha-1)}{2!}z^2 +  \frac{\alpha(\alpha-1)(\alpha-2)}{3!}z^3  
+ \cdots \right] \tag{**}\label{main-step}
\end{align}
\end{subequations}
%\end{eqnarray*}
在本题中，由 $f(0)=i-1$ 所确定的解析分支为
\begin{eqnarray*}
f(z) &=& \sqrt{z-2i} = \sqrt{-2i}\sqrt{1+\frac{z}{-2i}} = (i-1)\left( 1+ \frac{z}{-2i} \right) ^{\frac{1}{2}} \\ 
&=& (i-1) \left[ 1+ \frac{1}{2} \left(\frac{z}{-2i}\right)  + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}\left(\frac{z}{-2i}\right)^2 +  \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3!}\left(\frac{z}{-2i}\right)^3 + \cdots \right] \\ 
&=& (i-1)\left( 1 + \frac{i}{4}z + \frac{1}{32}z^2 - \frac{i}{128} z^3 + \cdots  \right) 
\end{eqnarray*}

注：一般幂函数的泰勒展式 (\ref{main-step}) 不容易从上一步（通过幂级数的指数函数）计算得到。
另一种方法是对 $e^{\alpha\mathrm{Ln}(1+z)}$ 先确定解析分支，然后使用复合函数求导方法，多次求导可以比较容易地得到这个泰勒展开。

}

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\item  %Problem 3 第123页例子4.14
设有泰勒展开 $ \frac{1}{1+z+z^2} = \sum\limits_{n=0}^{\infty} c_nz^n$. 
\begin{enumerate}
\item  求出收敛范围。
\item  求出系数 $c_n$ 的递推关系式。 
\item  求出展式的前五项。
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
函数 $f(z) = \frac{1}{1+z+z^2} = \frac{1}{(z-z_1)(z-z_2)}$, 其中 $z_{1,2} = \frac{-1\pm \sqrt{3}i }{2}$. 这两个根的模长都是1，
所以 $f(z)$ 在圆盘 $|z|<1$ 中是解析函数。根据泰勒定理，幂级数展开式
$ f(z) = \sum\limits_{n=0}^{\infty} c_nz^n$ 的系数可以通过柯西积分公式计算，取 $0<\rho<1$,  则
\begin{subequations}
\begin{align}
c_n = \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{f(\zeta)}{(\zeta-0)^{n+1}}d\zeta 
= \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{1}{\zeta^{n+1}(1+\zeta+\zeta^2)}d\zeta.  
\tag{*}
\label{taylor-1}
\end{align}
\end{subequations}
用 $n+1$ 和 $n+2$ 代替上式的 $n$, 可得
\begin{subequations}
\begin{align}
c_{n+1} =& \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{1}{\zeta^{n+2}(1+\zeta+\zeta^2)}d\zeta, \tag{**}\label{taylor-2}\\ 
c_{n+2} =& \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{1}{\zeta^{n+3}(1+\zeta+\zeta^2)}d\zeta. \tag{***}\label{taylor-3} 
\end{align}
\end{subequations}
观察 (\ref{taylor-1}),  (\ref{taylor-2}) 和 (\ref{taylor-3}) 可知，加起来可以简化积分，对任意 $n\ge 0$, 有 
\begin{subequations}
\begin{align}
c_n + c_{n+1} + c_{n+2} = \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{\zeta^2+\zeta+1}{\zeta^{n+3}(1+\zeta+\zeta^2)}d\zeta 
= \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{1}{\zeta^{n+3}}d\zeta 
= 0. 
\tag{$\heartsuit$}
\label{taylor-4}
\end{align}
\end{subequations}

因此所求的递推关系式为 $c_n = -c_{n-1}-c_{n-2}, \, n\ge 2$. 

注：通过这种方法，我们可以得出 $\frac{1}{1+az+bz^2}$ 的泰勒展开的系数之间的关系式。

为求出泰勒展式的前五项，我们只需要求出前两项，然后使用上述递推关系式。可以使用公式 (\ref{taylor-1}) 来计算。
也可以使用 $\frac{1}{1-t}$ 的展开，然后代入 $t=-z-z^2$. 
\begin{subequations}
\begin{align}
\frac{1}{1+z+z^2} =& \frac{1}{1-t} = 1+t+t^2+t^3+\cdots \nonumber \\ 
=& 1+ (-z-z^2) + (-z-z^2)^2 + (-z-z^2)^3 + \cdots \nonumber \\ 
=& 1- z +z^3 + \cdots. \nonumber 
%\tag{$\heartsuit$}
%\label{taylor-4}
\end{align}
\end{subequations}
因此 $c_0=1, c_1=-1$. 根据递推关系式 $c_n = -c_{n-1}-c_{n-2}$, 可得 $c_2=0$, $c_3=1$, $c_4=-1$, $c_5=0$, $c_6=1$. 
从而所求的前几项为 
\begin{subequations}
\begin{align}
\frac{1}{1+z+z^2} =  1 - z + z^3 - z^4 +z^6 - \cdots \nonumber 
%\tag{$\heartsuit$}
%\label{taylor-4}
\end{align}
\end{subequations}


}

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\item  %Problem 4 第页例子
将函数 $f(z)=\int_0^z \frac{\sin z}{z} dz$ 展成 $z$ 的幂级数，并指出展式成立的范围。

\vspace{0.2cm}

{\color{red}解答：
被积函数的泰勒展开为 
\begin{subequations}
\begin{align}
\frac{\sin(z)}{z} =  \frac{z-\frac{1}{3!}z^3+\frac{1}{5!}z^5-\frac{1}{7!}z^7+ }{z} \cdots 
=  1-\frac{1}{3!}z^2 + \frac{1}{5!}z^4 -\frac{1}{7!}z^6 +  \cdots, \nonumber 
%\tag{$\heartsuit$}
%\label{taylor-4}
\end{align}
\end{subequations}
该幂级数在整个复平面的每一点都收敛，因此在任意一个有界闭集上一致收敛。逐项积分可得
\begin{subequations}
\begin{align}
\int_0^z \frac{\sin(z)}{z}dz = \int_0^z\left(1-\frac{1}{3!}z^2 + \frac{1}{5!}z^4 -\frac{1}{7!}z^6 +  \cdots \right) 
= z-\frac{1}{3\cdot 3!}z^3 + \frac{1}{5\cdot 5!}z^5 -\frac{1}{7\cdot 7!}z^7 +  \cdots. \nonumber 
%\tag{$\heartsuit$}
%\label{taylor-4}
\end{align}
\end{subequations}
该展式在整个复平面上成立。
}

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\item  %Problem 5 第125页例子4.16
求 $f(z)=2-\sin(z)$ 的全部零点和阶数。

\vspace{0.2cm}

{\color{red}解答：
根据正弦函数的定义，转化成指数函数，从 $f(z)=0$ 可得
\begin{subequations}
\begin{align}
\frac{e^{iz}-e^{-iz}}{2i} =  2. \nonumber 
%\tag{$\heartsuit$}
%\label{taylor-4}
\end{align}
\end{subequations}
记 $e^{iz}=t$, 则有
\begin{eqnarray*}
t - \frac{1}{t} &=& 4i,  \\ 
t^2-4it-1 &=& 0,  \\ 
t &=& \frac{4i \pm \sqrt{-16+4}}{2} = 2i \pm \sqrt{3}i. 
\end{eqnarray*}
因此所求的零点为
\begin{eqnarray*}
z &=& \frac{1}{i}\mathrm{Ln}(2i\pm \sqrt{3}i) \\ 
&=&  -i\left( \ln (2\pm \sqrt{3}) +\ln(i) + 2k\pi i \right) \\ 
&=& -i \left( \ln (2\pm \sqrt{3}) +\frac{\pi}{2}i + 2k\pi i \right) \\ 
&=& \frac{\pi}{2} + 2k\pi -  \ln (2\pm \sqrt{3}) i, \,\, k=0, \pm 1, \pm 2, \cdots. 
\end{eqnarray*}

为计算这些零点的阶数，我们计算 $f'(z)$, 然后代入这些零点。
\begin{eqnarray*}
f'(z)\mid_{z=z_0} &=& -\cos(z)\mid_{z=z_0} \\ 
&=& - \cos\left( \frac{\pi}{2} + 2k\pi -  \ln (2\pm \sqrt{3}) i \right) \\ 
&=& - \cos \left(  \ln (2\pm \sqrt{3}) i \right)  \\ 
&=& - \frac{1}{2} \left[ e^{i[ \ln (2\pm \sqrt{3})i ]} + e^{-i[ \ln (2\pm \sqrt{3})i ] } \right] \\ 
&=& - \frac{1}{2} \left[ e^{- \ln (2\pm \sqrt{3}) ]} + e^{ \ln (2\pm \sqrt{3}) ] } \right] \\ 
&=& -\frac{1}{2} \left[ \frac{1}{2\pm \sqrt{3}} + (2\pm \sqrt{3})  \right] \\ 
&=&  -\frac{1}{2} \left[ 2\mp \sqrt{3} + (2\pm \sqrt{3})  \right] \\ 
&=& -2 \neq 0. 
\end{eqnarray*}
因此这些零点都是一阶的。

}

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\item  %Problem 6 第125页例子4.16
考虑函数 $f(z)=\left\{\begin{array}{ll} z^2\sin\frac{1}{z},& z\neq 0, \\ 0,& z=0. \end{array}\right. $
\begin{enumerate}
\item  求 $f(z)$ 的全部零点。
\item  函数 $f(z)$ 在 $z=0$ 是否解析？
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
首先由 $f(z)$ 的定义，可知 $z=0$ 是 $f(z)$ 的一个零点。其它零点形如 $\frac{1}{z_0}$, 其中 $\sin(z_0)=0$. 
由正弦函数的定义，$\sin(z)=(e^{iz}-e^{-iz})/(2i)$, 因此 $e^{iz_0} = e^{-iz_0}$, 即 $e^{2iz_0}=1$. 
因此 $2i z_0=2k\pi i$, 即 $z_0=k\pi$. 因此 $f(z)$ 的全部零点为 
$$0, \frac{1}{k\pi}, \,\, k=\pm 1, \pm 2, \cdots. $$
因为$z=0$ 是这些零点的聚点，但是 $f(z)$ 不是零函数，所以根据解析函数的零点的孤立性，可知 $f(z)$ 在 $z=0$ 不是解析的。

}

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\end{enumerate}


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